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\usepackage{amsmath} %支持更多数学格式
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\title{第二章   矩阵分解}
\author[]{曹金}

\institute{National Key Laboratory of Science and Technology on Communications at UESTC}
%\titlegraphic{\includegraphics[width=20mm]{USTL}}

\date{2012}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

\begin{frame}
  \frametitle{主要内容}
  \tableofcontents
\end{frame}


\section{三角分解}

\begin{frame}
  \frametitle{正线三角矩阵}
  \begin{block}{定义}
    如果$a_{ii} (i = 1, 2, \dots, n)$均为正实数, $a_{ij} \in C(R) ( i < j, i = 1, 2, \dots, n - 1; j = i + 1, i + 2, \dots, n)$, 则上三角矩阵
    \[
    R = \left[
    \begin{array}{cccc}
      a_{11} & a_{12} & \cdots & a_{1n} \\
      0  & a_{22} & \cdots & a_{2n} \\
      \vdots &  &  & \vdots \\
      0 & 0 & \cdots & a_{nn} \\
    \end{array} \right]
    \]
  \end{block}
  称为正线上三角复(实)矩阵. 特别当$a_{ii} = 1(i = 1, 2, \dots, n)$时, R称为单位上三角复(实)矩阵. \pause

  类似定义正线下三角矩阵.
\end{frame}

\begin{frame}
  \frametitle{矩阵的三角分解} \pause
  将矩阵分解为正线三角矩阵和酉矩阵的乘积. \pause
  \begin{block}{定理}
    设$A \in C_n^{n \times n}$, 则A可唯一地分解为\\
    \[
    A = U_1 R,
    \] \\ 
    或, \\
    \[
    A = L U_2,
    \]
  \end{block}
  其中$U_1, U_2$是酉矩阵, R和L分别是正线上三角复矩阵和正线下三角复矩阵. \pause\\
  证明.
\end{frame}


\section{谱分解}

%\subsection{单纯矩阵的谱分解}

\begin{frame}
  \frametitle{单纯矩阵}
  \begin{block}{定义} \pause
    如果矩阵A的几何重复度与代数重复度相等, 则称A为单纯矩阵.
  \end{block}
\end{frame}

\begin{frame}
  \frametitle{单纯矩阵的谱分解}
  将单纯矩阵分解成幂等矩阵的线性组合. \pause
  \begin{block}{定理}
    设$A \in C^{n \times n}$是单纯矩阵, 则A可分解为一系列幂等矩阵$A_i (i = 1,2,\dots,n)$的加权和, 即:
    \[
    A = \sum_{i = 1}^n \lambda_i A_i
    \]
    其中$\lambda_i (i = 1, 2, \dots, n)$是A的特征值.
  \end{block} \pause
  证明.
\end{frame}

\begin{frame}
  \frametitle{单纯矩阵的谱分解定理}
  更一般地, 单纯矩阵的谱分解定理为:
  \begin{block}{定理}
    设$A \in C^{n \times n}$, 它有k个相异特征值$\lambda_i (i = 1, 2, \dots, k)$, 则A是单纯矩阵的重要条件是存在k个矩阵$A_i (i = 1, 2, \dots, k)$满足:
    \begin{enumerate} \pause
    \item $
      A_i A_j = \left\{
        \begin{array}{rl}
          A_i & \quad j = i, \\
          O & \quad j \neq i;
        \end{array}
      \right.
      $
    \item $ \sum_{i = 1}^k A_k = E_n; $
    \item $ A = \sum_{i = 1}^k \lambda_i A_i $
    \end{enumerate}
  \end{block}
\end{frame}


%\subsection{正规矩阵的谱分解}

\begin{frame}
  \frametitle{正规矩阵}
  \begin{block}{定义}
    若n阶复矩阵A满足
    \[
    A A^H = A^H A
    \]
    则称A为正规矩阵. \pause\\
    当A为n阶实矩阵且满足
    \[
    A A^T = A^T A
    \]
    则称矩阵A为实正规矩阵.
  \end{block}
\end{frame}

\begin{frame}
  \frametitle{正规矩阵}
  \begin{block}{定理}
    n阶复矩阵A是正规矩阵的充要条件是A与对角阵酉相似. 即存在n阶酉矩阵U, 使得
    \[
    A = U diag(\lambda_i, \lambda_2, \dots, \lambda_n) U^H
    \]
    其中, $\lambda_i, \lambda_2, \dots, \lambda_n$是A的n个特征值.
  \end{block}
\end{frame}


\begin{frame}
  \frametitle{正规矩阵的谱分解}
  \begin{block}{定理} \pause
    设$A \in C^{n \times n}$, A有k个相异特征值$\lambda_i (i = 1, 2, \dots, k)$, 则A是正规矩阵的充要条件是存在k个矩阵$A_i (i = 1, 2, \dots, k)$使其满足:
    \begin{enumerate}
    \item $
      A_i A_j = \left\{
        \begin{array}{rl}
          A_i & \quad j = i, \\
          O & \quad j \neq i;
        \end{array}
      \right.
      $
    \item $ \sum_{i = 1}^k A_k = E_n; $
    \item $ A = \sum_{i = 1}^k \lambda_i A_i $
    \item $ A_i^H = A_i (i = 1, 2, \dots, k) $
    \end{enumerate}
  \end{block}
\end{frame}



\section{奇异值分解}


\begin{frame}
  \frametitle{奇异值分解} \pause
  \begin{block}{$A^H A$的性质}
    设$A \in C^{m \times n}$, 则有
    \begin{enumerate}
    \item $ rank(A)=rank(A^HA) = rank(AA^H) $
    \item $A^HA, AA^H$的特征值均为非负实数.
    \item $A^HA\text{与} AA^H$的非零特征值相同.
    \end{enumerate}
  \end{block}
\end{frame}


\begin{frame}
  \frametitle{奇异值分解} \pause
  \begin{block}{奇异值的定义}
    设$A \in C_r^{m \times n}$, $A^H A$的特征值为
    \[
    \lambda_1 \ge \lambda_2 \ge \cdots \ge \lambda_r \ge \lambda_{r + 1} = \cdots = \lambda_m = 0,
    \]
    则称$\sigma_i = \sqrt{\lambda_i} (i = 1, 2, \cdots, r)$为矩阵A的正奇异值.
  \end{block}
\end{frame}

\begin{frame}
  \frametitle{奇异值分解}
  \begin{block}{定理}
    设$A \in C_r^{m \times n}, \sigma_1, \sigma_2, \cdots, \sigma_r$是A的r个正奇异值, 则存在m阶酉矩阵U以及n阶酉矩阵V, 使得
    \[
    A = U \left(
      \begin{array}{cc}
        D & O \\
        O & O \\
      \end{array}
      \right) V
    \]
    其中, $D = diag (\delta_1, \delta_2, \cdots, \delta_r)$, 而$\delta_i$是满足$| \delta_i | = \sigma_i (i = 1, 2, \cdots, r)的复数$.
  \end{block}
\end{frame}

\begin{frame}
  \frametitle{证明} \pause
  因为$A^HA$是正规矩阵, 所以存在n阶酉矩阵V, 使得
  \[  
  A^HA = V^H diag(\sigma_1^2, \sigma_2^2, \dots, \sigma_r^2, 0, \dots, 0) V
  \] \pause
  \[
  V A^HA V^H = diag(\sigma_1^2, \sigma_2^2, \dots, \sigma_r^2, 0, \dots, 0)
  \] \pause
  设$V = \left( 
    \begin{array}{c}
      V_1 \\
      V_2 \\
    \end{array}
    \right)$, 其中$V_1 \in C_r^{r \times n}, V_2 \in C_{n - r}^{(n - r) \times n},$代入上式, 得:\pause
    \[
    VA^H A V^H = \left[
      \begin{array}{cc}
        V_1 A^H A V_1^H & V_1 A^H A V_2^H \\
        V_2 A^H A V_1^H & V_2 A^H A V_2^H \\
      \end{array}
    \right] =
    \left[
      \begin{array}{cc}
        D^H D & 0 \\
        0 & 0 \\
      \end{array}
    \right]
    \]
\end{frame}

\begin{frame}
    可以得到:
    \[ V_1 A^H A V_1^H = D^H D \]
    \[ V_2 A^H A V_2^H = O \] \pause
    \[ V_2 A^H A V_2^H = (A V_2^H)^H (A V_2^H)  = O \] \pause
  所以, $ A V_2^H  = O $ \pause \\
  令$U_1^H = (D^H)^{-1} V_1 A^H \in C^{r \times m}$, 有$U_1 = A V_1^H D^{-1}$. \\ \pause
  再将$U_1^H$扩充成m阶酉矩阵$U^H = \left(
    \begin{array}{c}
      U_1^H \\
      U_2^H \\
    \end{array}
    \right)$, 现在如果有
    \[
    U^H A V^H = \left(
      \begin{array}{c}
        V_1^H \\
        V_2^H \\
      \end{array}
    \right)  A (V_1^H, V_2^H) = \left[
      \begin{array}{cc}
        U_1^H A V_1^H & U_1^H A V_2^H \\
        U_2^H A V_1^H & U_2^H A V_2^H \\
      \end{array}
    \right] = \left(
      \begin{array}{cc}
        D & O \\
        O & O \\
      \end{array}
    \right)
    \]
    即能证明.
\end{frame}

\begin{frame}
    由U矩阵的定义可得:
    \[
    U_2^H U_1 = U_2^H A V_1^H D^{-1} = O
    \] \pause
    因为, D是A的正奇异值构成的对角阵, $D \neq O$, 所以有
    \[
    U_2^H A V_1^H = O
    \] \pause
    \[
    U_1^H A V_1^H = (D^H)^{-1} V_1 A^H A V_1^H = (D^H)^{-1}D^HD = D
    \]
    \[
     U_1^H A V_2^H = U_2^H A V_2^H = O,
    \]
\end{frame}


\begin{frame}
  \frametitle{奇异值分解在MIMO中的应用}
  \begin{block}{mimo信道模型}
    \begin{figure}
      \centering
      \includegraphics[width=0.5\textwidth]{mimo1.png}
      \caption{mimo模型}
    \end{figure}
    \[
    \mathbf{y} = H \mathbf{x} + \mathbf{w}, \quad H = 
    \left( 
      \begin{array}{cc}
        h_{11} & h_{21} \\
        h_{12} & h_{22} \\
      \end{array}
    \right)
    \] 
  \end{block}
\end{frame}

\begin{frame}
  \frametitle{SVD分解后的mimo信道模型}
  \begin{block}{}
    \begin{figure}
      \centering
      \includegraphics[width=0.9\textwidth]{mimo2.png}
      \caption{mimo模型}
    \end{figure}
    \[
    \tilde{y}_i = \lambda_i \tilde{x}_i + \tilde{w}_i, \quad i = 1,2,\dots,n_{min}. 
    \quad \text{设}
    \left\{\begin{array}{c}
        \tilde{x} := V^\ast x \\
        \tilde{y} := U^\ast y \\
        \tilde{w} := U^\ast w \\
      \end{array}
      \] 
    \end{block}
  \end{frame}

\section{最大秩分解}

\begin{frame}
  \frametitle{最大秩分解}
  \begin{block}{定理}
    设$A \in C_r^{m \times n}$, 则存在矩阵$B \in C_r^{m \times r}, D \in C_r^{r \times n}$, 使得
    \[
    A = BD
    \]
  \end{block} \pause
  注意: 最大秩分解不唯一. \pause \\
  证明.
\end{frame}

\begin{frame}
  \frametitle{最大秩分解定理}
  \begin{block}{}
    设$A \in C_r^{m \times n}$, 且$A = B_1 D_1 = B_2 D_2$均为A的最大秩分解, 则
    \begin{enumerate}
    \item 存在r阶可逆矩阵Q, 使得
      \[
      B_1 = B_2 Q, \quad D_1 = Q^{-1} D_2 ;
      \]
    \item $ D_1^H ( D_1 D_1^H )^{-1} (B_1^H B_1)^{-1} B_1^H = D_2^H ( D_2 D_2^H )^{-1} (B_2^H B_2)^{-1} B_2^H $
    \end{enumerate}
  \end{block}
\end{frame}



\begin{frame}
  \begin{block}{}
    问题讨论
  \end{block}
\end{frame}







\end{document}

